The only values found for p + q totaling 1015 or less are (2,2) and (2,3).
I have no proof that higher values of p+q won't lead to other solutions.
For any given total of p+q, the left side of the equation exceeds the right hand side when p and q are most different; the right side exceeds the left side when p and q are more nearly equal. Thus in the continuous case (with factorials replaced by a modified gamma function and integers by reals) there are many solutions--in fact two for any given value of p+q. To the point in this puzzle, I don't know if some of these solutions, at higher values of p+q, might happen to involve only integers.
10 for T=2 to 1015
20 for P=1 to T-1
30 Q=T-P
40 A=!(P)+!(Q)
50 B=P^Q
60 if A=B then print P,Q
70 next
80 next
(If P+Q = T were allowed to be 1016, the program overflows trying to take the factorial of 1015.)
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Posted by Charlie
on 2008-12-21 15:18:29 |
computer exploration
